In the diagram, the circle is inscribed in the square. This means that the circle and the square share points $S$, $T$, $U$, and $V$, and the width of the square is exactly equal to the diameter of the circle. Rounded to the nearest tenth, what is $XY/XV$, to the nearest tenth?
We see that $\angle XSY$ is a right angle using the radius and tangent at that point. Therefore, line $SY$ contains the diameter of the circle. By symmetry, we get $\angle XVY=\frac{\angle Y}{2}$.
It follows that $\angle XYV=\frac{180-\angle Y}{2}$ so $\angle XVY+\angle XYV=90$ and $\triangle XYV$ is a right triangle. Since $YV$ is a radius of the circle, it is half the length of the square's side and $XY/XV=\boxed{0.7}$.
To find $XY/XV$, we need to determine the lengths of $XY$ and $XV$.
Let's analyze the given information. We know that the circle is inscribed in the square, meaning that the width of the square is equal to the diameter of the circle.
Let's call the side length of the square $a$. Then, the diameter of the inscribed circle is also $a$.
Since the circle is inscribed in the square, the diameter (or side length) of the square is equal to the diagonal of the square.
Using the Pythagorean theorem, we can find the length of the diagonal.
The diagonal of a square with side length $a$ is equal to $a\sqrt{2}$.
Therefore, $XV = a\sqrt{2}$.
Now, let's consider triangle $XYS$.
Triangle $XYS$ is a right triangle, as the two radii that connect the center of the circle to the points of tangency with the square are perpendicular to the side of the square.
Since $XS$ is the radius of the circle, which is half of the diameter, and $XV$ is the diagonal of the square, we can find $XY$ using the Pythagorean theorem.
$XY^2 = XV^2 - XS^2$
$XY^2 = (a\sqrt{2})^2 - (a/2)^2$
$XY^2 = 2a^2 - a^2/4$
$XY^2 = 8a^2/4 - a^2/4$
$XY^2 = 7a^2/4$
Taking the square root of both sides, we find:
$XY = a\sqrt{\frac{7}{4}}$
Now, we can substitute the values of $XY$ and $XV$ into the expression $XY/XV$:
$\frac{XY}{XV} = \frac{a\sqrt{\frac{7}{4}}}{a\sqrt{2}}$
Simplifying, we find:
$\frac{XY}{XV} = \sqrt{\frac{7}{4}}/\sqrt{2}$
Rationalizing the denominator, we get:
$\frac{XY}{XV} = \sqrt{\frac{7}{4}}/\sqrt{2} \times \frac{\sqrt{2}}{\sqrt{2}}$
$\frac{XY}{XV} = \sqrt{\frac{7}{2 \cdot 4}}$
$\frac{XY}{XV} = \sqrt{\frac{7}{8}}$
$\frac{XY}{XV} = \frac{\sqrt{7}}{\sqrt{8}}$
$\frac{XY}{XV} = \frac{\sqrt{7}}{2\sqrt{2}}$
Since we are asked to round to the nearest tenth, we can approximate $\sqrt{7}$ and $\sqrt{2}$ to the nearest tenth:
$\sqrt{7} \approx 2.6$
$\sqrt{2} \approx 1.4$
Substituting these values, we find:
$\frac{XY}{XV} \approx \frac{2.6}{2 \cdot 1.4} \approx \frac{2.6}{2.8} \approx 0.93$
Therefore, rounded to the nearest tenth, $XY/XV \approx 0.9$.