Given $\mathbf{v}=4 \mathbf{i}-\mathbf{j}$ and $\mathbf{w}=3 \mathbf{i}+2 \mathbf{j}$, find the angle between $\mathbf{v}$ and $\mathbf{w}$.

The angle between $\mathbf{v}$ and $\mathbf{w}$ is $\square^{\circ}$.
(Do not round until the final answer. Then round to the nearest tenth as needed.)

Step 1 ：Given vectors \(\mathbf{v}=4 \mathbf{i}-\mathbf{j}\) and \(\mathbf{w}=3 \mathbf{i}+2 \mathbf{j}\), we are to find the angle between \(\mathbf{v}\) and \(\mathbf{w}\).

Step 2 ：First, we calculate the dot product of the vectors \(\mathbf{v}\) and \(\mathbf{w}\). The dot product of two vectors is calculated as the sum of the products of their corresponding components. In this case, \(\mathbf{v} \cdot \mathbf{w} = (4)(3) + (-1)(2) = 10\).

Step 3 ：Next, we calculate the magnitudes of the vectors \(\mathbf{v}\) and \(\mathbf{w}\). The magnitude of a vector is calculated as the square root of the sum of the squares of its components. In this case, \(||\mathbf{v}|| = \sqrt{(4)^2 + (-1)^2} = 4.123105625617661\) and \(||\mathbf{w}|| = \sqrt{(3)^2 + (2)^2} = 3.605551275463989\).

Step 4 ：Finally, we substitute these values into the formula for \(\theta\) and calculate the angle. The angle \(\theta\) is given by \(\theta = \cos^{-1}\left(\frac{\mathbf{v} \cdot \mathbf{w}}{||\mathbf{v}|| \cdot ||\mathbf{w}||}\right) = \cos^{-1}\left(\frac{10}{4.123105625617661 \cdot 3.605551275463989}\right) = 0.8329812666744316\) radians.

Step 5 ：We convert the angle to degrees to get \(\theta = 47.72631099390626\) degrees.

Step 6 ：Rounding to the nearest tenth, we get \(\theta = 47.7\) degrees.

Step 7 ：Final Answer: The angle between \(\mathbf{v}\) and \(\mathbf{w}\) is \(\boxed{47.7^{\circ}}\).