1. **State the problem:** We need to find the resultant vector of two given vectors and express it in rectangular coordinates rounded to 1 decimal place. 2. **Given:** - Vector 1: magnitude $3$, angle $70^\circ$ from positive x-axis (upwards and left means angle is $180^\circ - 70^\circ = 110^\circ$ from positive x-axis). - Vector 2: magnitude $6$, angle $35^\circ$ from positive x-axis (downwards and right means angle is $360^\circ - 35^\circ = 325^\circ$ from positive x-axis). 3. **Formula:** To convert a vector from polar to rectangular coordinates: $$x = r \cos \theta$$ $$y = r \sin \theta$$ 4. **Calculate components of Vector 1:** $$x_1 = 3 \cos 110^\circ$$ $$y_1 = 3 \sin 110^\circ$$ 5. **Calculate components of Vector 2:** $$x_2 = 6 \cos 325^\circ$$ $$y_2 = 6 \sin 325^\circ$$ 6. **Evaluate components using cosine and sine values:** $$x_1 = 3 \times \cos 110^\circ = 3 \times (-0.3420) = -1.026$$ $$y_1 = 3 \times \sin 110^\circ = 3 \times 0.9397 = 2.819$$ $$x_2 = 6 \times \cos 325^\circ = 6 \times 0.8192 = 4.915$$ $$y_2 = 6 \times \sin 325^\circ = 6 \times (-0.5736) = -3.442$$ 7. **Sum components to find resultant vector:** $$x_R = x_1 + x_2 = -1.026 + 4.915 = 3.889$$ $$y_R = y_1 + y_2 = 2.819 + (-3.442) = -0.623$$ 8. **Round to 1 decimal place:** $$x_R \approx 3.9$$ $$y_R \approx -0.6$$ **Final answer:** The resultant vector in rectangular coordinates is approximately $\boxed{(3.9, -0.6)}$.