Solve -1^4-6*(-1/3)^2+(-5)*(-3)

$$- 1 ^ { 4 } - 6 \times ( - \frac { 1 } { 3 } ) ^ { 2 } + ( - 5 ) \times ( - 3 )$$

Evaluate

$$\frac{40}{3}\approx 13.333333333$$

Solution Steps

Calculate 1 to the power of 4 and get 1.

$$-1-6\left(-\frac{1}{3}\right)^{2}-5\left(-3\right)$$

Calculate -\frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$-1-6\times \frac{1}{9}-5\left(-3\right)$$

Multiply 6 and \frac{1}{9} to get \frac{6}{9}.

$$-1-\frac{6}{9}-5\left(-3\right)$$

Reduce the fraction \frac{6}{9} to lowest terms by extracting and canceling out 3.

$$-1-\frac{2}{3}-5\left(-3\right)$$

Convert -1 to fraction -\frac{3}{3}.

$$-\frac{3}{3}-\frac{2}{3}-5\left(-3\right)$$

Since -\frac{3}{3} and \frac{2}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-3-2}{3}-5\left(-3\right)$$

Subtract 2 from -3 to get -5.

$$-\frac{5}{3}-5\left(-3\right)$$

Multiply -5 and -3 to get 15.

$$-\frac{5}{3}+15$$

Convert 15 to fraction \frac{45}{3}.

$$-\frac{5}{3}+\frac{45}{3}$$

Since -\frac{5}{3} and \frac{45}{3} have the same denominator, add them by adding their numerators.

$$\frac{-5+45}{3}$$

Add -5 and 45 to get 40.

$$\frac{40}{3}$$

Factor

$$\frac{2 ^ {3} \cdot 5}{3} = 13\frac{1}{3} = 13.333333333333334$$