Compute the dot product of the vectors \( \mathbf{v} \) and \( \mathbf{w} \), and find the angle between the vectors. \( \mathbf{v}=-9 \mathbf{i}-\mathbf{j} \) and \( \mathbf{w}=-\mathbf{i}-9 \mathbf{j} \) \( \mathbf{v} \cdot \mathbf{w}=18 \) (Simplify your answer.) Find the magnitude of the vectors. \( |\mathbf{v}|=\sqrt{82} \) and \( |\mathbf{w}|=\sqrt{82} \) (Type exact answers, using radicals as needed.) The angle between the vectors is \( \square^{\circ} \). (Type your answer in degrees. Do not round until the final answer. Then round to the nearest tenth as needed.

To find the angle between the vectors \( \mathbf{v} \) and \( \mathbf{w} \), we can use the dot product formula: \[ \mathbf{v} \cdot \mathbf{w} = |\mathbf{v}| |\mathbf{w}| \cos(\theta) \] Given: \[ \mathbf{v} \cdot \mathbf{w} = 18 \] \[ |\mathbf{v}| = \sqrt{82}, \quad |\mathbf{w}| = \sqrt{82} \] Plugging these values into the formula: \[ 18 = \sqrt{82} \times \sqrt{82} \times \cos(\theta) \] \[ 18 = 82 \cos(\theta) \] \[ \cos(\theta) = \frac{18}{82} = \frac{9}{41} \] Now, solve for \( \theta \): \[ \theta = \cos^{-1}\left(\frac{9}{41}\right) \] Using a calculator to find the inverse cosine: \[ \theta \approx \cos^{-1}(0.2195) \approx 77.4^\circ \] **Answer:** The angle between the vectors is \( \boxed{77.4^\circ} \).