1. **State the problem:** We need to find the length of the hypotenuse $x$ in a right triangle $MNO$ where the right angle is at $N$, the leg $NO$ is 7 units, and the angle at $O$ is 52°. 2. **Identify the sides relative to angle $O$:** The side $NO=7$ is adjacent to angle $O$, and $x=MN$ is the hypotenuse opposite the right angle. 3. **Use the cosine function:** For angle $O$, cosine relates the adjacent side and hypotenuse: $$\cos(52^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{7}{x}$$ 4. **Solve for $x$:** $$x = \frac{7}{\cos(52^\circ)}$$ 5. **Calculate the cosine value:** $$\cos(52^\circ) \approx 0.6157$$ 6. **Substitute and compute:** $$x = \frac{7}{0.6157}$$ 7. **Show cancellation step:** $$x = \frac{7}{\cancel{0.6157}} \times \frac{1}{\cancel{0.6157}} = 11.36$$ 8. **Round to the nearest tenth:** $$x \approx 11.4$$ **Final answer:** The length of the hypotenuse $x$ is approximately 11.4 units.