The volume of a large tank is \( 189 \mathrm{yd}^{3} \). It is \( 3 \frac{3}{5} \mathrm{yd} \) wide and \( 4 \frac{2}{3} \mathrm{yd} \) high. What is the length of the tank? Length \( =\square \) (Type an integer, fraction, or mixed number.)

To find the length of the tank, we can use the formula for the volume of a rectangular prism, which is given by: \[ \text{Volume} = \text{Length} \times \text{Width} \times \text{Height} \] We know the volume of the tank is \( 189 \mathrm{yd}^{3} \), the width is \( 3 \frac{3}{5} \mathrm{yd} \) (which is \( \frac{18}{5} \) in improper fraction form), and the height is \( 4 \frac{2}{3} \mathrm{yd} \) (which is \( \frac{14}{3} \) in improper fraction form). First, let's convert the width and height to improper fractions: - Width: \( 3 \frac{3}{5} = \frac{18}{5} \) - Height: \( 4 \frac{2}{3} = \frac{14}{3} \) Substituting these values into the volume formula: \[ 189 = \text{Length} \times \left(\frac{18}{5}\right) \times \left(\frac{14}{3}\right) \] Now we can multiply the width and height: \[ \frac{18}{5} \times \frac{14}{3} = \frac{18 \times 14}{5 \times 3} = \frac{252}{15} = \frac{84}{5} \] Now substituting back into the volume equation: \[ 189 = \text{Length} \times \frac{84}{5} \] To isolate the length, we multiply both sides by the reciprocal of \( \frac{84}{5} \): \[ \text{Length} = 189 \times \frac{5}{84} \] Calculating this gives: \[ \text{Length} = \frac{189 \times 5}{84} \] Now simplify \( \frac{189}{84} \): - \( 189 \div 21 = 9 \) - \( 84 \div 21 = 4 \) So, \( \frac{189}{84} = \frac{9}{4} \). Therefore: \[ \text{Length} = \frac{9}{4} \times 5 = \frac{45}{4} \text{ or } 11 \frac{1}{4} \mathrm{yd} \] So, in the answer box: Length \( = 11 \frac{1}{4} \)