Solve -3ab^2/2c^3div9a^2b/2c^2

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

Evaluate

$$-\frac{b}{3ac}$$

Solution Steps

Divide -\frac{3ab^{2}}{2c^{3}} by \frac{9a^{2}b}{2c^{2}} by multiplying -\frac{3ab^{2}}{2c^{3}} by the reciprocal of \frac{9a^{2}b}{2c^{2}}.

$$\frac{\left(-\frac{3ab^{2}}{2c^{3}}\right)\times 2c^{2}}{9a^{2}b}$$

Express \left(-\frac{3ab^{2}}{2c^{3}}\right)\times 2 as a single fraction.

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

Cancel out 2 in both numerator and denominator.

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

Express \frac{-3ab^{2}}{c^{3}}c^{2} as a single fraction.

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

Cancel out c^{2} in both numerator and denominator.

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

Express \frac{\frac{-3ab^{2}}{c}}{9a^{2}b} as a single fraction.

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

Cancel out 3ab in both numerator and denominator.

$$\frac{-b}{3ac}$$

Cancel out -1 in both numerator and denominator.

$$\frac{b}{-3ac}$$

Differentiate w.r.t. b

$$-\frac{1}{3ac}$$