Solve -1^4+(1-0.5)*1/3*lfloor2--3^2rfloor

$$- { 1 }^{ 4 } +(1-0.5) \times \frac{ 1 }{ 3 } \times \lfloor 2- { -3 }^{ 2 } \rfloor$$

Evaluate

$$-\frac{13}{6}\approx -2.166666667$$

Solution Steps

Calculate 1 to the power of 4 and get 1.

$$-1+\left(1-0.5\right)\times \frac{1}{3}\lfloor 2-\left(-3\right)^{2}\rfloor$$

Subtract 0.5 from 1 to get 0.5.

$$-1+0.5\times \frac{1}{3}\lfloor 2-\left(-3\right)^{2}\rfloor$$

Convert decimal number 0.5 to fraction \frac{5}{10}. Reduce the fraction \frac{5}{10} to lowest terms by extracting and canceling out 5.

$$-1+\frac{1}{2}\times \frac{1}{3}\lfloor 2-\left(-3\right)^{2}\rfloor$$

Multiply \frac{1}{2} times \frac{1}{3} by multiplying numerator times numerator and denominator times denominator.

$$-1+\frac{1\times 1}{2\times 3}\lfloor 2-\left(-3\right)^{2}\rfloor$$

Do the multiplications in the fraction \frac{1\times 1}{2\times 3}.

$$-1+\frac{1}{6}\lfloor 2-\left(-3\right)^{2}\rfloor$$

Calculate -3 to the power of 2 and get 9.

$$-1+\frac{1}{6}\lfloor 2-9\rfloor$$

Subtract 9 from 2 to get -7.

$$-1+\frac{1}{6}\lfloor -7\rfloor$$

The floor of a real number a is the largest integer number less than or equal to a. The floor of -7 is -7.

$$-1+\frac{1}{6}\left(-7\right)$$

Multiply \frac{1}{6} and -7 to get \frac{-7}{6}.

$$-1+\frac{-7}{6}$$

Fraction \frac{-7}{6} can be rewritten as -\frac{7}{6} by extracting the negative sign.

$$-1-\frac{7}{6}$$

Convert -1 to fraction -\frac{6}{6}.

$$-\frac{6}{6}-\frac{7}{6}$$

Since -\frac{6}{6} and \frac{7}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-6-7}{6}$$

Subtract 7 from -6 to get -13.

$$-\frac{13}{6}$$

Factor

$$-\frac{13}{6} = -2\frac{1}{6} = -2.1666666666666665$$