Solve -1^4+(1-05)*1/3*lfloor2--3^2rfloor

$$- { 1 }^{ 4 } +(1-05) \times \frac{ 1 }{ 3 } \times \lfloor 2- { -3 }^{ 2 } \rfloor$$

Evaluate

$$-\frac{10}{3}\approx -3.333333333$$

Solution Steps

Calculate 1 to the power of 4 and get 1.

$$-1+\left(1-0\times 5\right)\times \frac{1}{3}\lfloor 2-\left(-3\right)^{2}\rfloor$$

Multiply 0 and 5 to get 0.

$$-1+\left(1-0\right)\times \frac{1}{3}\lfloor 2-\left(-3\right)^{2}\rfloor$$

Subtract 0 from 1 to get 1.

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

Multiply 1 and \frac{1}{3} to get \frac{1}{3}.

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

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

$$-1+\frac{1}{3}\lfloor 2-9\rfloor$$

Subtract 9 from 2 to get -7.

$$-1+\frac{1}{3}\lfloor -7\rfloor$$

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

$$-1+\frac{1}{3}\left(-7\right)$$

Multiply \frac{1}{3} and -7 to get \frac{-7}{3}.

$$-1+\frac{-7}{3}$$

Fraction \frac{-7}{3} can be rewritten as -\frac{7}{3} by extracting the negative sign.

$$-1-\frac{7}{3}$$

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

$$-\frac{3}{3}-\frac{7}{3}$$

Since -\frac{3}{3} and \frac{7}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-3-7}{3}$$

Subtract 7 from -3 to get -10.

$$-\frac{10}{3}$$

Factor

$$-\frac{10}{3} = -3\frac{1}{3} = -3.3333333333333335$$