We are given the expression: \(\frac{22q^{2}}{9}\). Below are clear steps to rewrite and understand it.
- Rewrite the fraction: \(\frac{22q^{2}}{9}=\left(\frac{22}{9}\right)q^{2}\). This separates the numerical coefficient from the variable part.
- Factor the numerator: 22 = 2 × 11, so \(\frac{22}{9}=\frac{2\cdot11}{9}\). Since 2 and 11 share no common factors with 9, the fraction is in lowest terms (irreducible).
- Alternative forms:
- Decimal form: \(\frac{22}{9}\approx 2.444\ldots\), so the expression is approximately \(2.444\,q^{2}\).
- Mixed-number form: \(\frac{22}{9}=2\frac{4}{9}\), so the expression can be written as \(2\frac{4}{9}q^{2}\).
- Factor form: \(q^{2}=q\cdot q\), so the expression is also \(\frac{22}{9}\,q\cdot q\).
- Conditional simplification: If q is an integer multiple of 3 (say q = 3k), then q² = 9k² and the 9 in the denominator cancels: \(\frac{22q^{2}}{9}=\frac{22\cdot9k^{2}}{9}=22k^{2}\). In that special case the expression becomes an integer multiple of 22. Otherwise, the fraction does not simplify further.
- Final statement: The expression is already simplified as \(\frac{22}{9}q^{2}\). There are equivalent representations (decimal, mixed number), and only when q contains a factor of 3 does further cancellation occur.
Concise answer: \(\displaystyle \frac{22q^{2}}{9}=\frac{22}{9}q^{2}\) (irreducible unless q is a multiple of 3).