Solve (-b^2c/3a)*9a^2/2bc^2

$$( - \frac { b ^ { 2 } c } { 3 a } ) \cdot \frac { 9 a ^ { 2 } } { 2 b c ^ { 2 } }$$

Evaluate

$$-\frac{3ab}{2c}$$

Solution Steps

Multiply -\frac{b^{2}c}{3a} times \frac{9a^{2}}{2bc^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-b^{2}c\times 9a^{2}}{3a\times 2bc^{2}}$$

Cancel out 3abc in both numerator and denominator.

$$\frac{-3ab}{2c}$$

Differentiate w.r.t. c

$$\frac{3ab}{2c^{2}}$$