Calculate value of `(X+Y+Z)/(10)`, here X is O-N-O bond angle in `NO_(3)^(-)` Y is O-N-O bond angle in `NO_(2)^(+)` and Z is F-Xe-F adjacent bond angle in `XeF_(4)`.

Explanation

To solve the problem, we need to calculate the values of X, Y, and Z based on the bond angles in the given compounds and then find the value of \((X + Y + Z) / 10\).

### Step-by-Step Solution:

1. **Determine X (O-N-O bond angle in \(NO_3^-\))**:
- The structure of \(NO_3^-\) (nitrate ion) consists of one nitrogen atom centrally bonded to three oxygen atoms. The nitrogen is double-bonded to one oxygen and single-bonded to the other two, which carry a negative charge.
- The molecule exhibits resonance, where the double bond can be between nitrogen and any of the three oxygen atoms. This results in a planar structure.
- The bond angles in a planar structure with three equivalent bonds are \(120^\circ\).
- Therefore, \(X = 120^\circ\).

2. **Determine Y (O-N-O bond angle in \(NO_2^+\))**:
- The structure of \(NO_2^+\) (nitronium ion) consists of one nitrogen atom double-bonded to two oxygen atoms. There is a positive charge on the nitrogen.
- This molecule is also planar, and the arrangement of the atoms results in a linear configuration.
- The bond angle in a linear molecule is \(180^\circ\).
- Therefore, \(Y = 180^\circ\).

3. **Determine Z (F-Xe-F adjacent bond angle in \(XeF_4\))**:
- The structure of \(XeF_4\) (xenon tetrafluoride) has a central xenon atom bonded to four fluorine atoms in a square planar arrangement, with two lone pairs of electrons above and below the plane.
- The adjacent bond angle between the fluorine atoms is \(90^\circ\) due to the square planar geometry.
- Therefore, \(Z = 90^\circ\).

4. **Calculate \((X + Y + Z) / 10\)**:
- Now we can sum up the values of X, Y, and Z:
\[ X + Y + Z = 120 + 180 + 90 = 390 \]
- Now divide by 10:
\[ \frac{X + Y + Z}{10} = \frac{390}{10} = 39 \]

### Final Answer:
The value of \((X + Y + Z) / 10\) is \(39\).