Math Problem Solver

\(8.04\) rounded to the nearest \(\frac{1}{10}\) is \(8\).

Rounding \(8.04\) to the nearest \(\frac{1}{10}\) is equivalent to finding the integer multiple of \(\frac{1}{10}\) nearest to \(8.04\).

The two consecutive multiples of \(\frac{1}{10}\) that \(8.04\) falls between are:\[80 \cdot \frac{1}{10} = 8\] and \[(80 + 1) \cdot \frac{1}{10} = 8.1\] So:\[8 \le 8.04 \le 8.1\]

The distance from \(8.04\) to \(8\) is \(\left| 8.04 - 8 \right| = 0.04\).

The distance from \(8.04\) to \(8.1\) is \(\left| 8.04 - 8.1 \right| = 0.06\).

\(8.04\) is closer to \(8\) than it is to \(8.1\), so \(8.04\) rounded to the nearest \(\frac{1}{10}\) is \(8\).

\(80\) from the above statement about consecutive multiples of \(\frac{1}{10}\) can be computed from the formula for the greatest integer \(n\) such that \(n \cdot \frac{1}{10} \le 8.04\): \[ n = \operatorname{floor}\left(8.04 \div \frac{1}{10} \right) = \operatorname{floor}\left(80.4 \right) = 80\]