Eighth grade \( > \) C. 18 Multiply and divide powers: variable bases 85 P \[ \begin{array}{ll}\text { 4)) Simplify. Express your answer using positive exponents. } \\ \qquad \frac{6 t^{9} u^{3} v}{2 t^{7} u^{4} v^{2} \cdot t^{2} u^{3} v^{3}}\end{array} \]

To simplify the expression \(\frac{6 t^{9} u^{3} v}{2 t^{7} u^{4} v^{2} \cdot t^{2} u^{3} v^{3}}\), first simplify the coefficients: \[ \frac{6}{2} = 3. \] Now, let's focus on the variables in the denominator, starting with \(t\): \[ t^{7} \cdot t^{2} = t^{7+2} = t^{9}, \] so we have \[ \frac{t^{9}}{t^{9}} = t^{9 - 9} = t^{0} = 1. \] Moving on to \(u\): \[ u^{4} \cdot u^{3} = u^{4+3} = u^{7}, \] therefore, \[ \frac{u^{3}}{u^{7}} = u^{3-7} = u^{-4} = \frac{1}{u^{4}}. \] For \(v\): \[ v^{2} \cdot v^{3} = v^{2+3} = v^{5}, \] thus, \[ \frac{v}{v^{5}} = v^{1-5} = v^{-4} = \frac{1}{v^{4}}. \] Now, putting it all together, we have: \[ 3 \cdot 1 \cdot \frac{1}{u^{4}} \cdot \frac{1}{v^{4}} = \frac{3}{u^{4} v^{4}}. \] So, the simplified expression, using only positive exponents, is: \[ \frac{3}{u^{4} v^{4}}. \]